Timeline for Conjecture: The number of points modulo $p$ of certain elliptic curve is $p$ or $p+2$ for $p$ of form $p=27a^2+27a+7$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 20, 2017 at 10:54 | vote | accept | joro | ||
| Oct 19, 2017 at 11:27 | comment | added | joro | @MaxAlekseyev possible generalization: mathoverflow.net/questions/283853/… | |
| Oct 19, 2017 at 11:14 | comment | added | joro | Generalization of this: mathoverflow.net/questions/283853/… | |
| Oct 18, 2017 at 17:46 | answer | added | GH from MO | timeline score: 22 | |
| Oct 18, 2017 at 15:43 | comment | added | Chris Wuthrich | @joro Sure, but for proving the conjecture you don't need $k$. I agree with WhatsUp. Here more hints of how to solve it (for $k=1$). Theorem 4 in Chapter 18 of Ireland-Rosen tells you what $a_p$ is when $p\equiv 1\pmod{3}$ in terms of sixth residue symbols. But since $4D=8$ is a cube this reduces to $(2/p)$. Then you need to factor $p$. I think $\pi = (9a+4-\omega)/(\omega-1)$ is a prime factor in $\mathbb{Z}[\omega]$ with $\pi\equiv 2\pmod{3}$. Here $\omega^2+\omega+1=0$. Hopefully that helps proving the conjecture. | |
| Oct 18, 2017 at 14:37 | comment | added | WhatsUp | The curve has CM by $\mathbb{Z}[\sqrt[3]{1}]$, thus corresponds to some character. This conjecture, if true, can probably be proved easily. But I'm too lazy to work out the details... | |
| Oct 18, 2017 at 13:26 | comment | added | joro | The cryptographic point of this: mathoverflow.net/questions/283767/… | |
| Oct 18, 2017 at 13:26 | comment | added | joro | @ChrisWuthrich I know, but for cryptographic reasons I want the order to be $p$, so I need a twist. | |
| Oct 18, 2017 at 13:13 | comment | added | Chris Wuthrich | The quadratic twist by $k$ multiplies $a_p$ by $(k/p)$ so you may assume $k=1$ in your conjecture. | |
| Oct 18, 2017 at 12:16 | comment | added | joro | @MaxAlekseyev probably this can be generalized by not using $2k^3$ but something else. | |
| Oct 18, 2017 at 11:39 | comment | added | Max Alekseyev | We have $(2k^3/p) = (2k/p)$. Does $(2k^3/p)$ suggest a generalization? | |
| Oct 18, 2017 at 11:24 | history | edited | joro | CC BY-SA 3.0 |
Added sage session
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| Oct 18, 2017 at 10:54 | history | asked | joro | CC BY-SA 3.0 |